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Calculating Probabilities Given the Discrete Uniform and the Binomial Distribution Functions

Calculating Probabilities Given the Discrete Uniform and the Binomial Distribution Functions

We can calculate and interpret probabilities of random variables that assume either the uniform distribution or the binomial distribution.

The Discrete Uniform Distribution

As you will recall, under the uniform distribution, all possible outcomes have equal probabilities. A good example might be the throw of a die, in which case each of the outcomes, 1, 2, 3, 4, 5, and 6, has a 1/6 probability of occurrence.

Define the random variable X such that Xi = i,     where i = 1, 2, 3, …, k)

$$ P(X = x) = \cfrac {1}{k} $$

$$ \text{Mean} =\cfrac {(k + 1)}{2} $$

$$ \text{Variance} = \cfrac {(k^2 – 1)}{12} $$

Example: Uniform Distribution Function

A uniform distribution function has been defined as:

X = {1, 3, 5, 7, 9}

Calculate P(X = 5) and F(5).

Solution

$$ \begin{align*} P(X = x) & = \cfrac {1}{k} \\ K & = 5\\ \end{align*} $$

Therefore,

$$ P(X = 5) =\cfrac {1}{5} = 0.2 $$

F(5) is the cumulative probability of an outcome less than or equal to 5. Thus,

$$ \begin{align*} F(5) & = P(X = 1) + P(X = 3) + P(X = 5) \\ & = 0.2 + 0.2 + 0.2 \\ & = 0.6 \\ \end{align*} $$

The Binomial Distribution

The binomial distribution is a sequence of n Bernoulli trials where the outcome for every trial can be a success or a failure.

Suppose the probability of a success is θ:

$$ P\left( X=x \right) =\left( \begin{matrix} n \\ x \end{matrix} \right) { \theta }^{ x }\left( 1-\theta \right) ^{ n-x },x=0,1,2,…,n;0<\theta <1 $$

Where

$$ \left( \begin{matrix} n \\ x \end{matrix} \right) ={ \quad }^{ n }{ C }_{ x }=\frac { n! }{ \left( n-x \right) !x! } $$

$$ \text{Mean of X} = n \theta $$

$$ \text{Variance of X} = n \theta (1 – \theta )$$

Note: sometimes the probability of success can be denoted by “p” and (1-p) to denote the probability of failure.

Example: Binomial Distribution

The probability of surviving an attack by a certain disease is 60%. What is the probability that at least 11 out of a group of 12 people who catch the disease will survive?

Solution

The number of survivors is distributed binomially with parameters n = 12 and θ = 0.6:

$$ \begin{align*} P(X \ge 11) & = P(X = 11 \text{ or } 12) \\ &={ _{ 12 }{ C }_{ 11 }{ \ast 0.6 }^{ 11 }*{ 0.4 }^{ 1 }+_{ 12 }{ C }_{ 12 }{ \ast 0.6 }^{ 12 } } \\ &= \cfrac {12!}{(1!11!)} * 0.6^{11} * 0.4^1 + \cfrac {12!}{(0!12!)} *0.6^{12} \\ &= 12 * 0.003628 * 0.4 + 0.6^{12} \\ &= 0.01959 \end{align*} $$

Question

A bowl contains blue and orange balls. The probability of drawing a blue ball in any attempt is 0.5. Suppose you draw 5 balls from the bag, what is the probability that 3 of the 5 balls drawn are blue?

A. 0.3125

B. 0.03125

C. 0.25

Solution

The correct answer is A.

P(X = 3) = 5C3 * 0.53 * 0.52

= 10 * 0.03125

= 0.3125

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